Quant Question Of The Day: 171
Trigonometry
The number of roots between 0 and π of the equation 2sin2x + 1 = 3sinx equals
(A) 2
(B) 4
(C) 1
(D) 3
The number of roots between 0 and π of the equation 2sin2x + 1 = 3sinx equals
(A) 2 Correct Anwer : (D) 3 Solution: Now, 2sin2x + 1 = 3sinx 2sin2x – 3sinx + 1 = 0 (2sinx – 1)(sinx – 1) = 0 Option (4) is correct.
(B) 4
(C) 1
(D) 3
sinx = ½ or sinx = 1
x = π/6, π– π/6, x = π/2
New Question: Coordinate Geometry
good information
By Những dòng bồn cầu thông minh ... , 6 months ago
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